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3 years ago

In most cases, letters of reconmendation are required for addmission to?

Physics

1 answer:

kykrilka [37]3 years ago

8 0

Answer:

Get your recommenders to mention diverse achievements. ...

Help your recommenders with relevant info. ...

The letter should always include examples of things you did. ...

The letter should show how you improved over time. ...

The tone of the letter should not be too dry.

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Name the material used to transfer of charges from one body to other

Eduardwww [97]

Answer:

idk just needs points

Explanation:

i need 20 words so sorry ig

6 0

3 years ago

Suppose we take a 1 m long uniform bar and support it at the 21 cm mark. Hanging a 0.40 kg mass on the short end of the beam res

Ipatiy [6.2K]

Answer:

The mass of the beam is = 29 kg.

Explanation:

A beam with mass 40 kg is shown in figure. Point S is the support point. Point B is the middle point on the beam where mass of the beam acts.

Taking moment about Point S

40 × 21 = $M_{beam}$ × 29

$M_{beam}$ = 29 kg

Therefore the mass of the beam is = 29 kg.

4 0

3 years ago

The net external force on a golf cart is 395.2 n north. if the cart has a total mass of 259.1 kg, what is the cart's acceleratio

yKpoI14uk [10]

Applying the Newton Second Law of motion that is F=ma
We have F=395.2N (North) and m=259.1Kg

Putting these values in the equation

395.2 = 259.1 x a than
for a = 395.2/259.1
a = 1.525 m/sec 2

8 0

3 years ago

What is the correct answer?

dexar [7]

Answer:

length=2

Explanation:

You simply find the zeros of the quartic polynomial

The factored form is (x+1)(x-5)(x-2)=0

The zeros become -1,5 and 2

Dimension 5 is already given and the question says length>width and 2>-1

Therefore length=2

5 0

3 years ago

A metal rod has a length of 123. cm at 200°C. At what temperature will the length be 92.6 cm if the coefficient of linear expans

mestny [16]

Answer:

$\theta_{2} = 15400^0 C$

Explanation:

The formula for linear expansivity, $\alpha = \frac{l_{2} - l_{1} }{l_{1} ( \theta_{2} - \theta_{1} )}$

original length, l₁ = 123 cm = 1.23 m

final length, l₁ = 92.6 cm =0.926 m

original temperature, θ₁ = 200°C

Linear expansivity, α = 2 * 10⁻⁵ °C⁻¹

Putting these values into the formula:

$2 * 10^{-5} = \frac{1.23 - 0.926 }{l_{1} ( \theta_{2} -200 )}\\ \theta_{2} -200 = \frac{0.304}{2 * 10^{-5} } \\\theta_{2} = 15200 + 200\\\theta_{2} = 15400^0 C$

7 0

3 years ago